Question

On a toss of two dice. A throws a total of 5. Then the probability that he will throw another 5 before the throws 7 is

• A. $\frac{2}{45}$
• B. $\frac{2}{5}$
• C. $\frac{1}{81}$
• D. 1919

Answer 1

The correct option is B $\frac{2}{5}$

$5$ can be thrown in $4$ ways and $7$ in $6$ ways in a single thrown with a pair of dice. Hence no. of ways of throwing neither $5$ nor $7$ is $36-(4+6)=26$

. Prob. of throwing $5$ in a single throw with a pair of dice is $=\frac{4}{36}=\frac{1}{9}$

. Probability of throwing neither $5$ nor $7$

$=\frac{26}{36}=\frac{13}{18}$

Hence the reg. probability

$=\left(\frac{1}{9}\right)+\left(\frac{13}{18}\right)+\left(\frac{1}{9}\right)+{\left(\frac{13}{18}\right)}^2\left(\frac{1}{9}\right)+{\left(\frac{13}{18}\right)}^2\left(\frac{1}{9}\right)$

$=\frac{\frac{1}{9}}{1-\frac{13}{18}}=\frac{2}{5}$